What happens when you shuffle across a carpeted floor?

A 12-volt car battery uses about 660 amps. How much power does it use?

alexira [117]

We know, Power = Voltage * Current
P = 660 * 12 = 7920 watt

In short, Your Answer would be Option A

Hope this helps!

8 0

4 years ago

Read 2 more answers

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (the train must see the car advancing at a lower speed), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (anyone in fact) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (relative to the ground):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, we have to do all the same but using $v_{AT}=v_A+v_T$ (the train must see the car advancing at a faster speed), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago

A trumpet player, using her ear, hears 5 beats per second when she plays a note on her trumpet and simultaneously sounds a 440 H

german

<h2>Frequency of trumpet wire = 445 Hz</h2>

Explanation:

The given frequency of tuning fork is 440 Hz . It produces 5 beats with trumpet wire . That means , the frequency of wire can be 440 ± 5

It can be either 445 or 435

Now the length of wire is increased , by which its frequency decreases . Because frequency is inversely proportional to length of wire .

If we decrease the frequency in 435 , the difference between tuning fork frequency and wire frequency will become greater than 5 even . So it cannot produce 3 beats with it .

If we decrease frequency from 445 , it can become 443 Hz . It gives 3 beat with the tuning fork as given .

Thus the initial frequency of wire is 445 Hz

4 0

3 years ago

A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.

lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × $10^4$ newtons/coulomb

We have the magnitude of the load:

q = 6.4 × $10 ^{-19}$ coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × $10^{-2}$ meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × $10^{-19}$ )(6.5 × $10^4$ )(1.2 × $10^{-2}$ )

PE = 5.0 x $10^{-16}$ joules

None of the options shown is correct.

6 0

3 years ago

Now Abel and Kato use what they learned to answer the following problem. The initial speed of a tennis ball is 54 m/s and the la

34kurt

Answer:

h=19.4m

R=199.07 m

Explanation:

To solve this problem we use the parabolic motion equations:

We define:

$v_{i}$ total initial speed =54 $\frac{m}{s}$

$\alpha$ =angle that forms the total initial speed with the horizontal line= 21°

$v_{ix}$ :initial speed component in horizontal direction

= $v_{i} cos\alpha$ =54*cos 21= 50.41 m/s

$v_{iy}$ :initial speed component in vertical direction

= $v_{i} sin\alpha$ =54*sin21=19.35 m/s

$v_{x}$ :horizontal speed at any point on the parabolic path

$v_{y}$ : vertical speed at any point on the parabolic path

g= acceleration of gravity= 9,8 $\frac{m}{s^{2} }$

Equation of the speed of the football in the vertical direction :

$(v_{y} )^{2} =(v_{yi} )^{2} -2*g*y$ Equation (1)

Calculation of the maximum height(h)

The speed of the ball (vy) in the vertical direction gradually decreases until its value is zero when it reaches the maximum height.

We replace y=h, $v_{y} =0$ , $v_{iy} = 19.39\frac{m}{s}$ , $g=9.8\frac{m}{s^{2} }$ in the equation(1)

$0=19.35^{2} -2*9.8*h$

$h=\frac{19.35^{2} }{2*9.8}$

h=19.1 m

Calculating of the range (R)

Fórmula: $R=v_{ix} *t (m)$ Equation (2)

R is the maximum horizontal distance the ball reaches.

The time (t) for the ball to reach R is twice the time the ball spends to reach the maximum height. Then, we calculate the time $(t_{h} )$ ) when the ball reaches the maximum height